Answer
See below
Work Step by Step
1. Find eigenvalues:
$\left (3-\lambda \right )^3=0$
$\lambda_1=\lambda_2= \lambda_3=3$
Checking:
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-1-\lambda & 2 & 2 \\
-4 & 5-\lambda & 2\\
-4 & 2 & 5-\lambda
\end{bmatrix}=\begin{bmatrix}
-4 & 2 & 2 \\
-4 & 2 & 2\\
-4 & 2 & 2
\end{bmatrix}$
We obtain reduced row echelon form:
$\begin{bmatrix}
-4 & 2 & 2| 0 \\
-4 & 2 & 2| 0 \\
-4 & 2 & 2 | 0
\end{bmatrix} \approx \begin{bmatrix}
2 & -1 & -1 | 0 \\
0 & 0 & 0 | 0\\
0 & 0 & 0
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r,s$ be free variables.
$\vec{V}=r(1,2,0)+s(0,-1,1) \\
E_2=\{(1,2,0) +(0,-1,1)\}
\rightarrow dim(E_2)=2 \ne 3$
Hence, matrix $A$ is defective.
